# Union Probability for Device Reliability ## Problem Restatement A device has three independent components: - **A** fails with probability \( P(A) = .01 \) - **B** fails with probability \( P(B) = .03 \) - **C** fails with probability \( P(C) = .04 \) The **device fails if any one of A, B, or C fails** (union of failures). ### Question **What is the probability that the device works?** --- ## Step 1: Probability the Device Fails Since the device fails if *any* component fails, we use the **union of probabilities**: \[ P(\text{Device fails}) = P(A \cup B \cup C) \] For independent events, the probability that *none* fail is easier to compute first. --- ## Step 2: Probability All Components Work The probability that each component **works**: - \( P(A') = 1 - P(A) = .99 \) - \( P(B') = 1 - P(B) = .97 \) - \( P(C') = 1 - P(C) = .96 \) Since failures are independent, the probability that all work: \[ P(\text{Device works}) = P(A' \cap B' \cap C') = P(A') \times P(B') \times P(C') \] \[ = .99 \times .97 \times .96 \] \[ = .92208 \] --- ## Step 3: Final Answers - **Probability the device works:** \(\boxed{.92208}\) (or about **92.2%**) - **Probability the device fails:** \(1 - .92208 = .07792\) (or about **7.8%**) --- ## Summary Table | Event | Probability | |-------------------|-------------| | Device works | .92208 | | Device fails | .07792 | --- ## Visual Explanation  *Venn diagram showing union of failures for independent components A, B, and C. The region outside all circles is the probability the device works.*